ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutation,now he ask you to restore the premutation. Pair (i, j)(i < j) is considered as a reverse log if Ai > Aj is matched.Input In the first line there is the num…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 175    Accepted Submission(s): 74 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutat…

ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 758    Accepted Submission(s): 359 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each pr…

最近在线段树的世界里遨游,什么都能用线段树做,这不又一道权值线段树了么. ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1029    Accepted Submission(s): 528 Problem Description ZYB has a premutation P ,but he o…

ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 637    Accepted Submission(s): 301 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each pre…

题目简述:有一个全排列,一直每个前缀区间的逆序对数,还原这个排列. fi记录逆序对数,pi记录该位置数值,则k=fi-f(i-1)表示前i-1个数比pi大的数的个数,那么只要在剩余元素求出按大小顺序第i-k个数字即可. 线段树+二分搜索,线段树bit[i]记录i的在剩余元素的排名顺序. /******************************* Date : 2015-12-06 19:49:59 Author : WQJ (1225234825@qq.com) Link : http:/…

题意:给出序列前k项中的逆序对数,构造出这个序列. 分析:使用权值线段树来确定序列元素. 逆序对的数量肯定是递增的,从最后一个元素开始逆向统计,则\(a[i] - a[i-1]\)即位置i之前比位置i上的数大的个数.则该位置的数应该是序列中第\(i-a[i]+a[i-1]\)大的数,通过权值线段树查询全局第k大的数,并将其数量-1. 初始的权值线段树每个值的数量都是1. #include <bits/stdc++.h> #define lson rt<<1 #define rson…

设f_if​i​​是第ii个前缀的逆序对数,p_ip​i​​是第ii个位置上的数,则f_i-f_{i-1}f​i​​−f​i−1​​是ii前面比p_ip​i​​大的数的个数.我们考虑倒着做,当我们处理完ii后面的数,第ii个数就是剩下的数中第f_i-f_{i-1}+1f​i​​−f​i−1​​+1大的数,用线段树和树状数组可以轻松地求出当前第kk个是11的位置,复杂度O(N \log N)O(NlogN). 1 #define cn(i,p,q) for(int i=p;i<=q;i++) 2…

[HDU5592] ZYB's Premutation 题目大意:一个由\([1,n]\)组成的数列,但不知道具体排列,但给出每个前缀的逆序对数目,让你还原排列 Solution 创造一颗\([1,n]\)的权值线段树,初始权值都为\(1\),我们从后往前离线处理,每次拿到一个前缀的逆序对数\(p[i]\),说明在\(i\)上的这个数字\(a_i\)前面有\(sum=p[i]-p[i-1]\)个数字比它大,所以\(a_i\)是\(a_1,\cdots ,a_i\)中第\(i -sum\)大的数字…

ZYB's Premutation Accepts: 220 Submissions: 983 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) 问题描述 ZYBZYBZYB有一个排列PPP,但他只记得PPP中每个前缀区间的逆序对数,现在他要求你还原这个排列. (i,j)(i<j)(i,j)(i < j)(i,j)(i<j)被称为一对逆序对当且仅当Ai>AjA_i&g…